Energy and radius of the first Bohr orbit of | vedclass

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$(A)$ The energy of the $n^{th}$ orbit is given by $E_{n} = -2.18 	imes 10^{-18} 	imes \frac{Z^{2}}{n^{2}} \ J$. For $He^{+}$ $(Z=2, n=1)$: $E_{1} =

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$E_{n}(Li^{2+}) = -19.62 	imes 10^{-18} \ J; r_{n}(Li^{2+}) = 17.6 \ pm; E_{n}(He^{+}) = -8.72 	imes 10^{-18} \ J; r_{n}(He^{+}) = 26.4 \ pm$

Vedclass · Answer

$(A)$ The energy of the $n^{th}$ orbit is given by $E_{n} = -2.18 	imes 10^{-18} 	imes \frac{Z^{2}}{n^{2}} \ J$. For $He^{+}$ $(Z=2, n=1)$: $E_{1} = -2.18 	imes 10^{-18} 	imes \frac{2^{2}}{1^{2}} = -8.72 	imes 10^{-18} \ J$. For $Li^{2+}$ $(Z=3, n=1)$: $E_{1} = -2.18 	imes 10^{-18} 	imes \frac{3^{2}}{1^{2}} = -19.62 	imes 10^{-18} \ J$. The radius of the $n^{th}$ orbit is given by $r_{n} = 52.9 	imes \frac{n^{2}}{Z} \ pm$. For $He^{+}$ $(Z=2, n=1)$: $r_{1} = 52.9 	imes \frac{1^{2}}{2} = 26.45 \ pm \approx 26.4 \ pm$. For $Li^{2+}$ $(Z=3, n=1)$: $r_{1} = 52.9 	imes \frac{1^{2}}{3} = 17.63 \ pm \approx 17.6 \ pm$.

Energy and radius of the first Bohr orbit of $He^{+}$ and $Li^{2+}$ are: $[$Given $R_{H} = 2.18 \times 10^{-18} \ J, a_{0} = 52.9 \ pm$ $]$

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